If $A={\sin }^{2}x+{\cos }^{4}x$, then for all real $x$:

$A={\sin }^{2}x+(1-{\sin }^{2}x{)}^2$ Let $t={\sin }^{2}x$, where $t\in [0,1]$.$A=t+(1-t{)}^2=t+1-2t+t^2=t^2-t+1$.This is a parabola opening upwards.Vertex at $t=-(-1)/(2\cdot 1)=1/2$.Minimum value at $t=1/2$: $A_{min}=(1/2{)}^2-(1/2)+1=1/4-1/2+1=3/4$.Maximum value at endpoints $t=0$ or $t=1$: $A_{max}=0-0+1=1$.So, $\frac{3}{4}\le A\le 1$.